A community for students.
Here's the question you clicked on:
 0 viewing
Jemurray3
 4 years ago
Derivative of x^2:
Jemurray3
 4 years ago
Derivative of x^2:

This Question is Closed

Jemurray3
 4 years ago
Best ResponseYou've already chosen the best response.4dw:1323757204190:dw

Jemurray3
 4 years ago
Best ResponseYou've already chosen the best response.4dw:1323757255000:dw

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0y=x^2 dy/dx=2x we know derivative of x^n=nx^(n1)

Jemurray3
 4 years ago
Best ResponseYou've already chosen the best response.4I'm not asking :) hold on

Jemurray3
 4 years ago
Best ResponseYou've already chosen the best response.4That tangent line has an equation \[ y  x_0^2 = m(xx_0) \] for some m. Let x > x0. Then, since the points on the line are less than the points on the curve, \[ x^2 \geq m(xx_0) + x_0^2\] \[x^2x_0^2 = (x+x_0)(xx_0) \geq m(xx_0) \] and xx0 is positive, so \[x+x_0 \geq m\] Next, for x < x0, the points on the line are likewise less than the points of the curve, so we get the same equation but now xx0 < 0 so we obtain \[x+x_0 \leq m\] now, for the first one, since x must be greater than or equal to x0, we can say that \[x+x_0 \geq 2x_0 \geq m\] and Likewise for the second, since x must be less than x0, \[x + x_0 \leq 2x_0 \leq m\] thus \[2x_0 \leq m \leq 2x_0\] and thus \[m = 2x_0\] so the slope of the tangent line to the curve y=x^2 at some point x0 is equal to 2x0, no difference quotients or limits required. Ta da!

Jemurray3
 4 years ago
Best ResponseYou've already chosen the best response.4Don't judge me, I was bored and decided to try a derivative without a limit. Shush. :)

Mahnoor
 4 years ago
Best ResponseYou've already chosen the best response.1It's really interesting, actually.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.